Question

Find the area of the region bounded by the parabola $x^2=4y$ and the line $x=4y-2$.

Answer 1

Given equations of curves are

$x^2=4y$ .......(1)

and, $x=4y-2$ ......(2)

Equation 1 represents a parabola which is open upward having vertex (0,0) and equation 2 represents a straight line.

On putting the value of 4y from equation 1 in equation 2, we get,

$x=x^2-2$

$x^2-x-2=0$

$(x+1)(x-2)=0$

$x=-1,2$

When $x=-1$, then from equation 1, $y=\frac{1}{4}$

and when $x=2$, then from equation 1, $y=1$

Therefore, points of intersection of given curves are $(-1,\frac{1}{4})$ and $(2,1)$.

Therefore,

Required area = Area of shaded region BOAB

$=\underset{-1}{\overset{2}{\int }}\left[y\right(line)-y(parabola\left)\right]dx$

$=\underset{-1}{\overset{2}{\int }}\left[\right(\frac{x+2}{4})-\frac{x^2}{4}]dx$

$=\frac{1}{4}\underset{-1}{\overset{2}{\int }}(x+2-x^2)dx$

$=\frac{1}{4}\left[\right(2+4-\frac{8}{3})-(\frac{1}{2}-2+\frac{1}{3}\left)\right]$

$=\frac{1}{4}(6-\frac{8}{3}+2-\frac{5}{6})$

$=\frac{1}{4}.\frac{27}{6}=\frac{9}{8}sq.units$